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Below is how I solved a problem to determine the regression formula to predict maximal oxygen consumption (VO2max) from scores on a 12...
Below is how I solved a problem to "determine the regression formula" to predict maximal oxygen consumption (VO2max) from scores on a 12 min. run. The information given and how I solved for a and b are shown below.
The next question says, "Using the prediction formulas developed in problem 3, what is the predicted VO2max for a participant who ran 2,954 m in 12 min?"
If X stays the same (12 min) but the mean of X changes from 2,853 m to 2,954 m, wouldn't I have to also recalculate the standard deviation and the correlation (r)?
If so, how do I do that when I do not have N or the original data?
Information given
X (12 min run)
Mx (mean of X) = 2,853 m
sx (Standard Deviation of X) = 305 m
r (correlation) = .79
Ŷ = V·O 2max
My (mean of Y) = 52.6 ml·kg·min−1
sy (Standard Deviation of Y) = 6.3 ml·kg·min−1
Solving for b
b = r (sy/ sx)
r (correlation) = .79
sy (Standard Deviation of Y) = 6.3 ml·kg·min−1
sx (Standard Deviation of X) = 305 m
b=0.79 * (6.3/305)
b =
